An `AC` circuit consists of a `220Omega` resistance andn a `0.7 H` choke. Find the power obsorbed from `220 V` and `50 Hz` source connected in this circuti if the resistance and choke are joined
(a) In series
(b) in parallel.

Explanation

a) In series the impedance of the circuit is,
`Z= sqrt(R^(2)+omega^(2)L^(2))=sqrt(R^(2)+(2pivL)^(2))`
`=sqrt((220)^(2)+(2 xx 3.14 xx 50 xx0.7)^(2))=311 Omega`
`therefore I_(rms) = V_(rms)/Z= 220 / 311=0.707 A`

and from the impedance diagram drawn above,
`cosphi=R/Z = 220/311 = 0.707`
`therefore` The power abosorbed in the circuit, `P=V_(rms)I_(rms)cosphi`
`=(220)(0.707)(0.707)`W
=110.08 W
b) When the resistance and choke are in parallel, the entire power is absorbed in resistance, as the choke having zero resistance). absorbs no power.
`therefore P=(V_(rms)^(2))/R=(220)^(2)/220 = 220V`